Problem: $A=\left[\begin{array}{rr}12 & 14 & -3 & 6 \\2 & 18 & -10 &12 \\1 &-9 &40 & -2 \\3 &3 &6 & 1\end{array}\right]$ $A_{2,4}=$
Solution: Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{2,4}$ $A_{{2},{4}}$ is located on row ${2}$ of $A$ : $\left[\begin{array}{rr}12 & 14 & -3 & 6 \\ {2} & {18} & {-10} & {12} \\1 &-9 &40 & -2 \\3 &3 &6 & 1\end{array}\right]$ $A_{{2},{4}}$ is also located on column ${4}$ of $A$. $\left[\begin{array}{rr}12 & 14 & -3 & 6 \\ {2} & {18} & {-10} & {\text{12}} \\1 &-9 &40 & {-2} \\3 &3 &6 & 1\end{array}\right]$ Therefore, $A_{{2},{4}}={12}$. Summary $A_{2,4}=12$